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prisoha [69]
3 years ago
15

Round to the nearest hundredth 5/8

Mathematics
1 answer:
Solnce55 [7]3 years ago
3 0
Eighths and their multiples are common fractions which I recommend memorizing, but to actually solve this, you use the literal meaning of a fraction and divide 5 by 8. See the long-division below (it was surprisingly difficult to type, so I hope it helps!).

To round 0.625 to the nearest hundredth, we go to the second decimal place, which is 5, so we round up to 0.63.

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6x-21>3 and 14x+11>-17
Delicious77 [7]
6x-21>3
Add 21 to both sides
6x>24
Divide 6 on both sides
X>4

14x+11>-17
Subtract 11 from both sides
14x>-28
Divide 14 on both sides
X<-2
8 0
3 years ago
Find pb, if pj=14 and jb= 28
anyanavicka [17]

Answer:

42

Step-by-step explanation:

Based on the information given, point j is the midpoint, as it is shared by both line segments. Set the equation:

pb = pj + jb

Note:

pj = 14

jb = 28

Plug in the corresponding numbers to the corresponding variables:

pb = pj + jb

pb = 14 + 28

pb = 42

pb = 42 is your answer.

~

3 0
3 years ago
Read 2 more answers
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
Please help! Its one of my last! Will give brainliest.
mixas84 [53]

Answer:

hexagon

Step-by-step explanation:

A hexagon has six vertices. Hope that this helps you and have a great day

7 0
3 years ago
Read 2 more answers
Concept of Function - Quiz - Level H
Hatshy [7]

Answer:

Step-by-step explanation:

It is the first box

6 0
3 years ago
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